SOLUTION

Average Voltage to Peak Voltage

The relationship between average voltage ($V_{avg}$) and peak voltage ($V_p$) for a half-wave rectified sine wave is given by:

$V_{avg} = \frac{V_p}{\pi}$

Rearranging gives:

$V_p = V_{avg} \times \pi$

Given $V_{avg} = 40V$:

$V_p = 40 \times \pi \approx 125.66V$

Calculating RMS Voltage

The RMS voltage ($V_{rms}$) for a half-wave rectified sine wave is calculated as:

$V_{rms} = \frac{V_p}{2}$

Substituting the value of $V_p$:

$V_{rms} = \frac{40\pi}{2} = 20\pi \approx 62.83V$

Calculating Power Dissipated in the Resistor

The power P dissipated in a resistor is given by:

$P = \frac{V_{rms}^2}{R}$

Substituting the value of V_rms and resistance R = 40Ω:

$P = \frac{(20\pi)^2}{40}$

Simplifying further:

$P = \frac{400\pi^2}{40} = 10\pi^2$

Decimal Approximation of Power

Using the approximation π²≈ 9.8696 we can calculate:

$P \approx 10 \times 9.8696 \approx 98.696W$

Therefore, the power dissipated in the resistor is approximately 98.69 Watts.